Question

Решите логарифмическое уравнение

Answer 1

Log [7] (6+7^(-x))=1+x=log [7] (7*7^(x))
6+7^(-x)=7*7^(x)
7^x=y
6+1/y=7*y
7y^2-6y-1=0
d=36+28=64
y1=(6+8)/14=1
y1=(6-8)/14=-1/7 - посторонний корень
7^x=y=1
х=0 - это ответ

Answer 2

Solving Logarithmic Equations

To solve logarithmic equations, we need to isolate the logarithm on one side of the equation and then apply the appropriate logarithmic properties to simplify and solve for the variable. Let's solve the logarithmic equations you provided one by one.

Equation 1: \(log_3(x^2+4)=log_3(4x)\)

To solve this equation, we can use the property that if two logarithms with the same base are equal, then their arguments must be equal. Therefore, we have:

\(x^2+4=4x\)

Rearranging the equation:

\(x^2-4x+4=0\)

This is a quadratic equation. Factoring or using the quadratic formula, we find that the solutions are:

\(x=2\) (multiplicity 2)

Equation 2: \(log_2(8)=log_{16}(x)+2\)

To solve this equation, we can use the property that \(log_a(b^c)=c \cdot log_a(b)\). Applying this property, we have:

\(3=log_{16}(x)+2\)

Subtracting 2 from both sides:

\(1=log_{16}(x)\)

Converting to exponential form:

\(16^1=x\)

Therefore, the solution is:

\(x=16\)

Equation 3: \(log_2(2x^2)-5=log_2(x) +log_2(x-5)\)

To solve this equation, we can use logarithmic properties to simplify the equation. First, we can combine the logarithms on the right-hand side using the property \(log_a(b) + log_a(c) = log_a(b \cdot c)\):

\(log_2(2x^2)-5=log_2(x(x-5))\)

Next, we can use the property \(log_a(b^c)=c \cdot log_a(b)\) to simplify the equation further:

\(log_2(2x^2)-5=log_2(x^2-5x)\)

Now, we can use the property that if two logarithms with the same base are equal, then their arguments must be equal:

\(2x^2-5=x^2-5x\)

Rearranging the equation:

\(x^2+5x-5=0\)

This is a quadratic equation. Factoring or using the quadratic formula, we find that the solutions are:

\(x \approx -6.854\) and \(x \approx 0.854\)

Please note that these solutions are approximate values rounded to three decimal places.

I hope this helps! Let me know if you have any further questions.